# Solve trigonometric function $x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0$

I need to solve a trigonometric function similar to the following one for $\alpha$.

$$x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0$$

I found a solution to a very similar problem

$$\frac{1}{2} \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0.$$

here, but the $\cos(2\alpha)$ term causes me some trouble, to adapt this solution.

TIA for your time!

## 1 Answer

Using the complex representation, we have $z:=e^{i\alpha}$, $\cos\alpha=\frac{z+z^{-1}}2$, $\sin\alpha=\frac{z-z^{-1}}{2i}$, and $\cos2\alpha=\frac{z^2+z^{-2}}2$, $\sin2\alpha=\frac{z^2-z^{-2}}{2i}$. $$x_1\frac{z^2-z^{-2}}{2i}+x_2\frac{z^2+z^{-2}}2-x_3\frac{z-z^{-1}}{2i}-x_4\frac{z+z^{-1}}2=0,$$ or $$x_1\frac{z^4-1}{2i}+x_2\frac{z^4+1}2-x_3\frac{z^3-z}{2i}-x_4\frac{z^3+z}2=0.$$

This is a general quartic equation in $z$ and unless the coefficients have special values, there is no easy solution.